\newproblem{lay:4_4_25}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 4.4.25}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Show that the subset $\{\mathbf{u}_1,\mathbf{u}_2,...,\mathbf{u}_p\}$ in $V$ is linearly independent if and only if the set of coordinate vectors
	$\{[\mathbf{u}_1]_B,[\mathbf{u}_2]_B,...,[\mathbf{u}_p]_B\}$ is linearly independent in $\mathbb{R}^n$. \textit{Hint}: Since the coordinate mapping is one-to-one,
	the following equations have the same solution $c_1,c_2,...,c_p$:
	\begin{center}
		$c_1\mathbf{u}_1+c_2\mathbf{u}_2+...+c_p\mathbf{u}_p=\mathbf{0}_V$ \\
		$c_1[\mathbf{u}_1]_B+c_2[\mathbf{u}_2]_B+...+c_p[\mathbf{u}_p]_B=\mathbf{0}$
	\end{center}
}{
  % Solution
	Consider the equation
	\begin{center}
		$c_1\mathbf{u}_1+c_2\mathbf{u}_2+...+c_p\mathbf{u}_p=\mathbf{0}_V$ \\
	\end{center}
	and its solution coefficients $c_1,c_2,...,c_p$. Since the coordinate mapping is one-to-one, the set of solutions is the same as the one for the equation
	\begin{center}
		$[c_1\mathbf{u}_1+c_2\mathbf{u}_2+...+c_p\mathbf{u}_p]_B=[\mathbf{0}_V]_B$ \\
	\end{center}
	and because the coordinate mapping is a linear transformation, we have
	\begin{center}
		$c_1[\mathbf{u}_1]_B+c_2[\mathbf{u}_2]_B+...+c_p[\mathbf{u}_p]_B=\mathbf{0}$
	\end{center}
	But, by hypothesis, the set $\{[\mathbf{u}_1]_B,[\mathbf{u}_2]_B,...,[\mathbf{u}_p]_B\}$ is linearly independent. That means that the only solution of the equation
	is $c_1=c_2=...=c_p=0$ and so is the solution of the first equation in the sequel.
}
\useproblem{lay:4_4_25}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}
